two physics a2 questions

mik1w

1. how do you test that microwaves diffract (5 marks)
this is an actual exam question and I missed the lesson when we did the experiment and I can't ask my teacher at the moment (weekend!)

we need microwave transmitter,two slits separation,a microwave detector ached to an amplifire
now move the the amplifire in a arc shape you'll find maxima's places and minima's places,this indicates that waves from the transmitter had diffracted and overlapped constructively and destructively
!)

2. a capacitor, battery and resistor are in series in a circuit. the capacitor starts wth no charge. how does the voltage across the capacitor vary with the charge on the capacitor? is it linear (and decreasing)? like I thought.
can someone explain why this is... I also have a hunch that the gradient is related to the capacitance but I can't quite sort it out in my head.?

Q is directley proportional to V
hence Q=cV
so the larger the Q th larger V
it's a stright line through the origin and no it's increasing the gradient of Q against V is the capacitance

also, how would you indicate the total energy stored on the capacitor ?

you find the voltage across the capacitor and as the capacitance of the capacitor is know usally factories write the value on it and you use
0.5CV^2=E

Reply 3

nas7232

for the first question, i had an exam which said draw an experiment where you can see if microwaves defract.

I put the young's double slit experiment with a single slit before the double slit. You cant see the interference pattern on the screen but you can use a detector. You will get a central maxima and a series of order maxima & minima on the detector. Not sure if that helps or not.
I haven't done phy5 yet so i can't really help with the 2nd question.

that is soo wrong
you can't spot a microwave on a screen

Reply 4

Christophicus

18

habosh

that is soo wrong
you can't spot a microwave on a screen

He said that. Read what he wrote again.

Reply 5

Widowmaker

He said that. Read what he wrote again.

I guess I wasn't actully concentration while reading :biggrin:

i have a concentration while reading problem problem which causes me to lose some marks in the exams

Reply 6

mik1w

2. a capacitor, battery and resistor are in series in a circuit. the capacitor starts wth no charge. how does the voltage across the capacitor vary with the charge on the capacitor? is it linear (and decreasing)? like I thought.
can someone explain why this is... I also have a hunch that the gradient is related to the capacitance but I can't quite sort it out in my head.

also, how would you indicate the total energy stored on the capacitor?

In a Q-V graph, the gradient is the capacitance.
gradient, m, = ΔQ/ΔV
=> C=Q/V

Energy stored is the area under a Q-V graph.
W = 0.5QV

Reply 7

OP

okay thanks that answeres both my questions.

for the energy represented on the graph, would that be the area under the straight line then as E = 0.5VQ. the graph now makes sense and seems kinda obvious but I needed soemone to jog my memory.

also what slit seperation will you use for the microwaves because their wavelength is much larger I think so the seperation should be bigger than with visible light? (fringe spacing = distance*slit spacing/wavelength? If i remember correctly)

thanks again for help reps to both of u if i remember to do it tomorrow

Reply 8

mik1w

okay thanks that answeres both my questions.

for the energy represented on the graph, would that be the area under the straight line then as E = 0.5VQ. the graph now makes sense and seems kinda obvious but I needed soemone to jog my memory.

Absolutely, here for a previous capacitors thread http://www.thestudentroom.co.uk/t96904.html

Reply 9

nas7232

19

i put the single slit before the double slit so that he makes the microwave emitter have coherant waves after the double slit screen. Is this right? As the microwave emitter may not be emitting coherant light.

Central maxima => path difference= 0, waves superpose constructively
minima => waves superpose deconstructively

Include:
order 1 maxima
order 2 maxima

Basically i wrote it in as much detail as possible it was 4 marks not 5 marks like your question. Hope that's right.

:tsr:

Reply 10

mik1w

also what slit seperation will you use for the microwaves because their wavelength is much larger I think so the seperation should be bigger than with visible light? (fringe spacing = distance*slit spacing/wavelength? If i remember correctly)

thanks again for help reps to both of u if i remember to do it tomorrow

I believe for this, as has been suggested, you have a transmitter behind a double slit, and a detector infront of the double slit. Move the detector, finding reinforcements and cancellations. When you find a cancellation, close one slit, and the signal should return as all the waves diffract through the other slit.

I think that's about right..

Reply 11

nas7232

19

mik1w

also what slit seperation will you use for the microwaves because their wavelength is much larger I think so the seperation should be bigger than with visible light? (fringe spacing = distance*slit spacing/wavelength? If i remember correctly)

thanks again for help reps to both of u if i remember to do it tomorrow

That would be good to write. I wish i thought of that on friday :eek:

.

Lambda = xs / D

I would write:
Make the slit seperation xMetres (can't remember how many, i think its under 10 cm [guess]), in order to get a good diffraction pattern.

Reply 12

OP

oops my equation was wrong.
i should remember it the two large figures (lambda and D) on one side and the two small ones (x and s) on the other: λD = xs

Reply 13

I read once slit separation should be from 5-7 cm as it has a large wavelength and diffraction is best when appreature size is comparable to wavelength

Reply 14

OP

i guess you could calculate knowing the ratios of the wavelengths and the slit seperation of red light

come to think of it, I don't know what it is for red light.. it is half a millimetre?

Reply 15

microwave superposition.JPG22.9KB

mik1w

i guess you could calculate knowing the ratios of the wavelengths and the slit seperation of red light

come to think of it, I don't know what it is for red light.. it is half a millimetre?

actully in an essay question I once have done they asked what is the slit separation,and you don't have anything to calculate in the question for red light the wavelength is larger than 700nm so it has to be extreamly small,but i don't think it's required is it?? oh I've just read it in theNAS book it's something close to .14mm,
BTW here is a poorly drowing of the experiment

Reply 16

mik1w

i guess you could calculate knowing the ratios of the wavelengths and the slit seperation of red light

come to think of it, I don't know what it is for red light.. it is half a millimetre?

Wavelength of red light = 650nm = 6.5*10^(-7)m
Assume you want fringe of 5mm = 0.005m; and distance = 5m.

Unparseable latex formula:

\large \lambda \ = \ ax/D \\ \, \\[br]a \ = \ \lambda D/x \\ \, \\[br]a \ = \ (5*6.5*10^{-7})/0.005 \\ \, \\[br]a \ = \ 6.5*10^{-4} m [br][br]

ie a = 0.65 mm

Reply 17

endeavour

Wavelength of red light = 650nm = 6.5*10^(-7)m
Assume you want fringe of 5mm = 0.005m; and distance = 5m.

Unparseable latex formula:

\large \lambda \ = \ ax/D \\ \, \\[br]a \ = \ \lambda D/x \\ \, \\[br]a \ = \ (5*6.5*10^{-7})/0.005 \\ \, \\[br]a \ = \ 6.5*10^{-4} m [br][br]

ie a = 0.65 mm

do you have a link to a page where I could find all wavelengths I need for A2 if you get what I mean

Reply 18

habosh

do you have a link to a page where I could find all wavelengths I need for A2 if you get what I mean

Quick google for electromagnetic spectrum yielded http://hyperphysics.phy-astr.gsu.edu/hbase/ems1.html

Reply 19

OP